Find the characteristic frequencies of a circular membrane which satisfies the Klein Gordon equation (Problem 25).

Klein-Gordon equation is ${\nabla }^{2}u=\left(\frac{1}{v^2}\right)\frac{{\partial }^{2}u}{\partial t^2}+{\lambda }^{2}u$.

Any disturbance or energy transfer from one location to another is referred to as a wave. Each wave is guided by a mathematical formula. A wave can be either standing or stationary.

Consider the Klein Gordon equation.

${\nabla }^{2}u=\left(\frac{1}{v^2}\right)\frac{{\partial }^{2}u}{\partial t^2}+{\lambda }^{2}u$

Express it in polar coordinates.

${\nabla }^{2}u=\frac{{\partial }^{2}u}{\partial r^2}+\frac{1}{r}\left(\frac{\partial u}{\partial r}\right)+\frac{1}{r^2}\left(\frac{{\partial }^{2}u}{\partial {\theta }^2}\right)$ ….. (1)

Separate the spatial and temporal variables to write the above equation in the form of $u=F\left(r,\theta \right)T\left(t\right)$.

Assume the separation constant.$-k^2$

$\left(\frac{{\partial }^{2}F}{\partial r^2}\right)+\frac{1}{r}\left(\frac{\partial F}{\partial r}\right)+\frac{1}{r^2}\left(\frac{{\partial }^{2}F}{\partial {\theta }^2}\right)+\left(k^2-{\lambda }^2\right)F=0$ ….. (2)

$\frac{d^{2}T}{d{t}^2}+k^2{v}^{2}T=0$ ….. (3)

Solve equation (3) to get $T=\left\{\begin{array}{l}Cos\left(kvt\right)\\ Sin\left(kvt\right)\end{array}\right.$.

Separate spatial variable from $F\left(r,\theta \right)$and express in terms of $F\left(r,\theta \right)=R\left(r\right)\Theta \left(\theta \right)$.

$\frac{1}{\Theta }\left(\frac{{\partial }^2\Theta }{\partial {\theta }^2}\right)=-n^2$

Solve the above equation to get$\Theta =\left\{\begin{array}{l}\cos \left(n\theta \right)\\ sin\left(n\theta \right)\end{array}\right.$ .

Write equation (2) for r.

$r^2\left(\frac{{\partial }^{2}R}{\partial r^2}\right)+r\left(\frac{\partial R}{\partial r}\right)-m^{2}R+\left(k^2-{\lambda }^2\right)r^{2}R=0$

Solve the equation by changing the variables $z={\left(k^2-{\lambda }^2\right)}^{\frac{1}{2}}r$.

This gives us the Bessel’s differential equation and the solution can be written in the form of $R=Z_p\left(z\right)$.

Change to the previous variable.

$R=Z_p\left(\sqrt{k^2-{\lambda }^{2}r}\right)$

Consider the requirement of finite u at r=0.

The general solution is$u=J_p\left(\sqrt{k^2-{\lambda }^{2}r}\right)\left(\begin{array}{c}\cos \left(n\theta \right)\\ sin\left(n\theta \right)\end{array}\right)\left(\begin{array}{c}\cos \left(kvt\right)\\ \sin \left(kvt\right)\end{array}\right)$ .

Consider the requirement of u=0.

At r=R the equation is:

$\sqrt{k^2-{\lambda }^2}R=k_{nm}$ ….. (4)

Square both sides in equation (4) and solve for $k^2$.

$k^2={\left(\frac{k_{nm}}{R}\right)}^2+{\lambda }^2$ ….. (5)

As $J_p\left(k_{nm}\right)=0$, $k_{nm}$is a solution of $J_p\left(z\right)$.

Replace R=a in equation (5) and find the characteristic frequencies of oscillation.

$$\begin{gathered} \nu =\frac{\omega }{2\pi } \\ =\frac{kv}{2\pi } \\ =\frac{v}{2\pi }\sqrt{{\left(\frac{k_{nm}}{a}\right)}^2+{\lambda }^2} \end{gathered}$$

Hence the characteristic frequencies of oscillation is $v=\frac{v}{2\pi }\sqrt{{\left(\frac{k_{nm}}{a}\right)}^2+{\lambda }^2}$.